354549
The figure shows two snap shots, each of a wave travelling along a particular string. If phase corresponds to snap shot 1 is \(4 x-8 t\) then phase corresponds to snap shot 2 is
1 \(4 x-8 t\)
2 \(8 x-16 t\)
3 \(4 x-4 t\)
4 \(4 x-16 t\)
Explanation:
\(y=f(k x-\omega t)\) For \(1, k=\dfrac{2 \pi}{\lambda}=4\) and \(\omega=8\) The wavelength of wave \( - 2\) is \(\dfrac{\lambda}{2}\) and frequency is double the initial frequency since \(v\) is same for both For \(2, k^{\prime}=\dfrac{2 \pi}{(\lambda / 2)}=4\) and \(\omega^{\prime}=2 \times 8=16\) \(\therefore\) phase of \(2=(8 x-16 t)\)
PHXI15:WAVES
354550
Equation of a wave is \(y=0.4 m \sin (3.14 t-3.14 x)\); where \(x\) is in metre and \(t\) is in second. Wavelength of the wave is
1 \(2\,m\)
2 \(0.5\,m\)
3 \(50\,m\)
4 \(25\,m\)
Explanation:
Comparing the given equation with \(\begin{aligned}& y=A \sin \left(2 \pi f t+\dfrac{2 \pi}{\lambda} x\right) \\& \dfrac{2 \pi}{\lambda}=3.14 \\& \therefore \lambda=2 m\end{aligned}\)
PHXI15:WAVES
354551
A progressive wave is represented by \(y = 12\sin (5t - 4x)cm\). On this wave, how far away are the two points having phase difference of \(90^{\circ}\) ?
1 \(\frac{\pi }{2}\;cm\)
2 \(\frac{\pi }{4}\;cm\)
3 \(\frac{\pi }{8}\;cm\)
4 \(\frac{\pi }{{16}}\;cm\)
Explanation:
According to question, the progressive wave is represented by \(y = 12\sin (5t - 4x)cm\) Comparing this equation with standard equation of progressive wave \(y=A \sin (\omega t-k x)\) So, we have \(A = 12\) \(\omega=5\) \(k = 4\) Here, \((\omega t-k x)\) is phase difference \(=\dfrac{\pi}{2}\) \(\therefore 5 t-4 x=\dfrac{\pi}{2}\) When \(t = 0,4x = \frac{\pi }{2}\) \(\therefore x = \frac{\pi }{8}\;cm\)
PHXI15:WAVES
354552
A wave is represented by \(y = 3\sin 2\pi \left( {\frac{t}{{0.04}} - \frac{X}{{0.01}}} \right)cm\). The frequency of the wave and the maximum acceleration under this frequency are
354553
A wave travelling along the \(x\)-axis is described by the equation \(y(x, t)=0.005 \cos (\alpha x-\beta t)\). If the wavelength and the time period of the wave are \(0.08\;m\) and \(2.0\;s\), respectively, then \(\alpha\) and \(\beta\) in appropriate units are
\(y(x, t)=0.005 \cos (\alpha x-\beta t)\) (Given) Comparing it with the standard equation of wave \(\begin{aligned}& y(x, t)=a \cos (k x-\omega t) \text { we get } \\& k=\alpha \text { and } \omega=\beta\end{aligned}\) But \(k=\dfrac{2 \pi}{\lambda}\) and \(\omega=\dfrac{2 \pi}{T}\) \(\Rightarrow \dfrac{2 \pi}{\lambda}=\alpha \text { and } \dfrac{2 \pi}{T}=\beta\) Given that \(\lambda = 0.08\;m\) and \(T = 2.0\;s\) \(\therefore \alpha=\dfrac{2 \pi}{0.08}=25 \pi \text { and } \beta=\dfrac{2 \pi}{2}=\pi\)
354549
The figure shows two snap shots, each of a wave travelling along a particular string. If phase corresponds to snap shot 1 is \(4 x-8 t\) then phase corresponds to snap shot 2 is
1 \(4 x-8 t\)
2 \(8 x-16 t\)
3 \(4 x-4 t\)
4 \(4 x-16 t\)
Explanation:
\(y=f(k x-\omega t)\) For \(1, k=\dfrac{2 \pi}{\lambda}=4\) and \(\omega=8\) The wavelength of wave \( - 2\) is \(\dfrac{\lambda}{2}\) and frequency is double the initial frequency since \(v\) is same for both For \(2, k^{\prime}=\dfrac{2 \pi}{(\lambda / 2)}=4\) and \(\omega^{\prime}=2 \times 8=16\) \(\therefore\) phase of \(2=(8 x-16 t)\)
PHXI15:WAVES
354550
Equation of a wave is \(y=0.4 m \sin (3.14 t-3.14 x)\); where \(x\) is in metre and \(t\) is in second. Wavelength of the wave is
1 \(2\,m\)
2 \(0.5\,m\)
3 \(50\,m\)
4 \(25\,m\)
Explanation:
Comparing the given equation with \(\begin{aligned}& y=A \sin \left(2 \pi f t+\dfrac{2 \pi}{\lambda} x\right) \\& \dfrac{2 \pi}{\lambda}=3.14 \\& \therefore \lambda=2 m\end{aligned}\)
PHXI15:WAVES
354551
A progressive wave is represented by \(y = 12\sin (5t - 4x)cm\). On this wave, how far away are the two points having phase difference of \(90^{\circ}\) ?
1 \(\frac{\pi }{2}\;cm\)
2 \(\frac{\pi }{4}\;cm\)
3 \(\frac{\pi }{8}\;cm\)
4 \(\frac{\pi }{{16}}\;cm\)
Explanation:
According to question, the progressive wave is represented by \(y = 12\sin (5t - 4x)cm\) Comparing this equation with standard equation of progressive wave \(y=A \sin (\omega t-k x)\) So, we have \(A = 12\) \(\omega=5\) \(k = 4\) Here, \((\omega t-k x)\) is phase difference \(=\dfrac{\pi}{2}\) \(\therefore 5 t-4 x=\dfrac{\pi}{2}\) When \(t = 0,4x = \frac{\pi }{2}\) \(\therefore x = \frac{\pi }{8}\;cm\)
PHXI15:WAVES
354552
A wave is represented by \(y = 3\sin 2\pi \left( {\frac{t}{{0.04}} - \frac{X}{{0.01}}} \right)cm\). The frequency of the wave and the maximum acceleration under this frequency are
354553
A wave travelling along the \(x\)-axis is described by the equation \(y(x, t)=0.005 \cos (\alpha x-\beta t)\). If the wavelength and the time period of the wave are \(0.08\;m\) and \(2.0\;s\), respectively, then \(\alpha\) and \(\beta\) in appropriate units are
\(y(x, t)=0.005 \cos (\alpha x-\beta t)\) (Given) Comparing it with the standard equation of wave \(\begin{aligned}& y(x, t)=a \cos (k x-\omega t) \text { we get } \\& k=\alpha \text { and } \omega=\beta\end{aligned}\) But \(k=\dfrac{2 \pi}{\lambda}\) and \(\omega=\dfrac{2 \pi}{T}\) \(\Rightarrow \dfrac{2 \pi}{\lambda}=\alpha \text { and } \dfrac{2 \pi}{T}=\beta\) Given that \(\lambda = 0.08\;m\) and \(T = 2.0\;s\) \(\therefore \alpha=\dfrac{2 \pi}{0.08}=25 \pi \text { and } \beta=\dfrac{2 \pi}{2}=\pi\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
354549
The figure shows two snap shots, each of a wave travelling along a particular string. If phase corresponds to snap shot 1 is \(4 x-8 t\) then phase corresponds to snap shot 2 is
1 \(4 x-8 t\)
2 \(8 x-16 t\)
3 \(4 x-4 t\)
4 \(4 x-16 t\)
Explanation:
\(y=f(k x-\omega t)\) For \(1, k=\dfrac{2 \pi}{\lambda}=4\) and \(\omega=8\) The wavelength of wave \( - 2\) is \(\dfrac{\lambda}{2}\) and frequency is double the initial frequency since \(v\) is same for both For \(2, k^{\prime}=\dfrac{2 \pi}{(\lambda / 2)}=4\) and \(\omega^{\prime}=2 \times 8=16\) \(\therefore\) phase of \(2=(8 x-16 t)\)
PHXI15:WAVES
354550
Equation of a wave is \(y=0.4 m \sin (3.14 t-3.14 x)\); where \(x\) is in metre and \(t\) is in second. Wavelength of the wave is
1 \(2\,m\)
2 \(0.5\,m\)
3 \(50\,m\)
4 \(25\,m\)
Explanation:
Comparing the given equation with \(\begin{aligned}& y=A \sin \left(2 \pi f t+\dfrac{2 \pi}{\lambda} x\right) \\& \dfrac{2 \pi}{\lambda}=3.14 \\& \therefore \lambda=2 m\end{aligned}\)
PHXI15:WAVES
354551
A progressive wave is represented by \(y = 12\sin (5t - 4x)cm\). On this wave, how far away are the two points having phase difference of \(90^{\circ}\) ?
1 \(\frac{\pi }{2}\;cm\)
2 \(\frac{\pi }{4}\;cm\)
3 \(\frac{\pi }{8}\;cm\)
4 \(\frac{\pi }{{16}}\;cm\)
Explanation:
According to question, the progressive wave is represented by \(y = 12\sin (5t - 4x)cm\) Comparing this equation with standard equation of progressive wave \(y=A \sin (\omega t-k x)\) So, we have \(A = 12\) \(\omega=5\) \(k = 4\) Here, \((\omega t-k x)\) is phase difference \(=\dfrac{\pi}{2}\) \(\therefore 5 t-4 x=\dfrac{\pi}{2}\) When \(t = 0,4x = \frac{\pi }{2}\) \(\therefore x = \frac{\pi }{8}\;cm\)
PHXI15:WAVES
354552
A wave is represented by \(y = 3\sin 2\pi \left( {\frac{t}{{0.04}} - \frac{X}{{0.01}}} \right)cm\). The frequency of the wave and the maximum acceleration under this frequency are
354553
A wave travelling along the \(x\)-axis is described by the equation \(y(x, t)=0.005 \cos (\alpha x-\beta t)\). If the wavelength and the time period of the wave are \(0.08\;m\) and \(2.0\;s\), respectively, then \(\alpha\) and \(\beta\) in appropriate units are
\(y(x, t)=0.005 \cos (\alpha x-\beta t)\) (Given) Comparing it with the standard equation of wave \(\begin{aligned}& y(x, t)=a \cos (k x-\omega t) \text { we get } \\& k=\alpha \text { and } \omega=\beta\end{aligned}\) But \(k=\dfrac{2 \pi}{\lambda}\) and \(\omega=\dfrac{2 \pi}{T}\) \(\Rightarrow \dfrac{2 \pi}{\lambda}=\alpha \text { and } \dfrac{2 \pi}{T}=\beta\) Given that \(\lambda = 0.08\;m\) and \(T = 2.0\;s\) \(\therefore \alpha=\dfrac{2 \pi}{0.08}=25 \pi \text { and } \beta=\dfrac{2 \pi}{2}=\pi\)
354549
The figure shows two snap shots, each of a wave travelling along a particular string. If phase corresponds to snap shot 1 is \(4 x-8 t\) then phase corresponds to snap shot 2 is
1 \(4 x-8 t\)
2 \(8 x-16 t\)
3 \(4 x-4 t\)
4 \(4 x-16 t\)
Explanation:
\(y=f(k x-\omega t)\) For \(1, k=\dfrac{2 \pi}{\lambda}=4\) and \(\omega=8\) The wavelength of wave \( - 2\) is \(\dfrac{\lambda}{2}\) and frequency is double the initial frequency since \(v\) is same for both For \(2, k^{\prime}=\dfrac{2 \pi}{(\lambda / 2)}=4\) and \(\omega^{\prime}=2 \times 8=16\) \(\therefore\) phase of \(2=(8 x-16 t)\)
PHXI15:WAVES
354550
Equation of a wave is \(y=0.4 m \sin (3.14 t-3.14 x)\); where \(x\) is in metre and \(t\) is in second. Wavelength of the wave is
1 \(2\,m\)
2 \(0.5\,m\)
3 \(50\,m\)
4 \(25\,m\)
Explanation:
Comparing the given equation with \(\begin{aligned}& y=A \sin \left(2 \pi f t+\dfrac{2 \pi}{\lambda} x\right) \\& \dfrac{2 \pi}{\lambda}=3.14 \\& \therefore \lambda=2 m\end{aligned}\)
PHXI15:WAVES
354551
A progressive wave is represented by \(y = 12\sin (5t - 4x)cm\). On this wave, how far away are the two points having phase difference of \(90^{\circ}\) ?
1 \(\frac{\pi }{2}\;cm\)
2 \(\frac{\pi }{4}\;cm\)
3 \(\frac{\pi }{8}\;cm\)
4 \(\frac{\pi }{{16}}\;cm\)
Explanation:
According to question, the progressive wave is represented by \(y = 12\sin (5t - 4x)cm\) Comparing this equation with standard equation of progressive wave \(y=A \sin (\omega t-k x)\) So, we have \(A = 12\) \(\omega=5\) \(k = 4\) Here, \((\omega t-k x)\) is phase difference \(=\dfrac{\pi}{2}\) \(\therefore 5 t-4 x=\dfrac{\pi}{2}\) When \(t = 0,4x = \frac{\pi }{2}\) \(\therefore x = \frac{\pi }{8}\;cm\)
PHXI15:WAVES
354552
A wave is represented by \(y = 3\sin 2\pi \left( {\frac{t}{{0.04}} - \frac{X}{{0.01}}} \right)cm\). The frequency of the wave and the maximum acceleration under this frequency are
354553
A wave travelling along the \(x\)-axis is described by the equation \(y(x, t)=0.005 \cos (\alpha x-\beta t)\). If the wavelength and the time period of the wave are \(0.08\;m\) and \(2.0\;s\), respectively, then \(\alpha\) and \(\beta\) in appropriate units are
\(y(x, t)=0.005 \cos (\alpha x-\beta t)\) (Given) Comparing it with the standard equation of wave \(\begin{aligned}& y(x, t)=a \cos (k x-\omega t) \text { we get } \\& k=\alpha \text { and } \omega=\beta\end{aligned}\) But \(k=\dfrac{2 \pi}{\lambda}\) and \(\omega=\dfrac{2 \pi}{T}\) \(\Rightarrow \dfrac{2 \pi}{\lambda}=\alpha \text { and } \dfrac{2 \pi}{T}=\beta\) Given that \(\lambda = 0.08\;m\) and \(T = 2.0\;s\) \(\therefore \alpha=\dfrac{2 \pi}{0.08}=25 \pi \text { and } \beta=\dfrac{2 \pi}{2}=\pi\)
354549
The figure shows two snap shots, each of a wave travelling along a particular string. If phase corresponds to snap shot 1 is \(4 x-8 t\) then phase corresponds to snap shot 2 is
1 \(4 x-8 t\)
2 \(8 x-16 t\)
3 \(4 x-4 t\)
4 \(4 x-16 t\)
Explanation:
\(y=f(k x-\omega t)\) For \(1, k=\dfrac{2 \pi}{\lambda}=4\) and \(\omega=8\) The wavelength of wave \( - 2\) is \(\dfrac{\lambda}{2}\) and frequency is double the initial frequency since \(v\) is same for both For \(2, k^{\prime}=\dfrac{2 \pi}{(\lambda / 2)}=4\) and \(\omega^{\prime}=2 \times 8=16\) \(\therefore\) phase of \(2=(8 x-16 t)\)
PHXI15:WAVES
354550
Equation of a wave is \(y=0.4 m \sin (3.14 t-3.14 x)\); where \(x\) is in metre and \(t\) is in second. Wavelength of the wave is
1 \(2\,m\)
2 \(0.5\,m\)
3 \(50\,m\)
4 \(25\,m\)
Explanation:
Comparing the given equation with \(\begin{aligned}& y=A \sin \left(2 \pi f t+\dfrac{2 \pi}{\lambda} x\right) \\& \dfrac{2 \pi}{\lambda}=3.14 \\& \therefore \lambda=2 m\end{aligned}\)
PHXI15:WAVES
354551
A progressive wave is represented by \(y = 12\sin (5t - 4x)cm\). On this wave, how far away are the two points having phase difference of \(90^{\circ}\) ?
1 \(\frac{\pi }{2}\;cm\)
2 \(\frac{\pi }{4}\;cm\)
3 \(\frac{\pi }{8}\;cm\)
4 \(\frac{\pi }{{16}}\;cm\)
Explanation:
According to question, the progressive wave is represented by \(y = 12\sin (5t - 4x)cm\) Comparing this equation with standard equation of progressive wave \(y=A \sin (\omega t-k x)\) So, we have \(A = 12\) \(\omega=5\) \(k = 4\) Here, \((\omega t-k x)\) is phase difference \(=\dfrac{\pi}{2}\) \(\therefore 5 t-4 x=\dfrac{\pi}{2}\) When \(t = 0,4x = \frac{\pi }{2}\) \(\therefore x = \frac{\pi }{8}\;cm\)
PHXI15:WAVES
354552
A wave is represented by \(y = 3\sin 2\pi \left( {\frac{t}{{0.04}} - \frac{X}{{0.01}}} \right)cm\). The frequency of the wave and the maximum acceleration under this frequency are
354553
A wave travelling along the \(x\)-axis is described by the equation \(y(x, t)=0.005 \cos (\alpha x-\beta t)\). If the wavelength and the time period of the wave are \(0.08\;m\) and \(2.0\;s\), respectively, then \(\alpha\) and \(\beta\) in appropriate units are
\(y(x, t)=0.005 \cos (\alpha x-\beta t)\) (Given) Comparing it with the standard equation of wave \(\begin{aligned}& y(x, t)=a \cos (k x-\omega t) \text { we get } \\& k=\alpha \text { and } \omega=\beta\end{aligned}\) But \(k=\dfrac{2 \pi}{\lambda}\) and \(\omega=\dfrac{2 \pi}{T}\) \(\Rightarrow \dfrac{2 \pi}{\lambda}=\alpha \text { and } \dfrac{2 \pi}{T}=\beta\) Given that \(\lambda = 0.08\;m\) and \(T = 2.0\;s\) \(\therefore \alpha=\dfrac{2 \pi}{0.08}=25 \pi \text { and } \beta=\dfrac{2 \pi}{2}=\pi\)
